On the level ground, the angle of elevation of the top of the tower is 30degrees, on moving 20 metres nearer; the angle of elevation is 60degrees. Then the height of the tower is

• First draw the right angle triangles according to the given condition.
  From rt angle triangle,
  tan60=h/x (h=height of the tower and x= dist between bottom of tower to 2nd
  root3=h/x point where the angle of elevation is 60 degree)
  also, tan30=h/(20+x)
  1/root3=h/(20+x)
  From above two equations, x*root3=(20+x)/root3
  3x=20+x
  so, x=10
  Then, h=x*root3=17.32m
• 12 years agoHelpfull: Yes(2) No(0)
• draw triangle BCA BCIS HYPOTHESIS BA AS PERPENDICULAR AND CA AS BASE
  NOW AB/AD=tan60
  AD=AB/ROOT3=h/ROOT3
  AB/AC=tan30
  AC=AB*ROOT3=h*ROOT3
  CD=(AC-AD)
  (h*ROOT3-h/ROOT3)=24
  h=12*root3
  h=20.76m
• 12 years agoHelpfull: Yes(1) No(2)
• draw a rt triangle ABC in which
• 12 years agoHelpfull: Yes(0) No(5)
• First of all draw a right angle triangle ABC such that BC=hypotenuse, AB= Height and AC be the base.
  Sol. Let AB be the tower and C and D be the points of observations. Then,
  AB/AD=tan60®= √3 .
  AD= AB/√3 =h/√3 .
  
  AB/AC =tan 30®=1/√3 , AC=AB*√3 =h√3 .
  CD= (AC-AD)=(h√3 -h/√3 ).
  h√3 -h/√3=20.
  h =10√3 =(10*1.73)=17.32
  hence the height of tower is 17.32 meter.
  
  
• 12 years agoHelpfull: Yes(0) No(0)